1. H.G/1100/061 Factoring a Polynomial and Rational Expressions Lecture #3 Dr.Hayk Melikyan Departmen of Mathematics and CS melikyan@nccu.edu

2. H.G/1100/062 Factoring is the process of writing a polynomial as the product of two or more polynomials. The factors of 6x 2 – x – 2 are 2x + 1 and 3x – 2. In this section, we will be factoring over the integers. Polynomials that cannot be factored using integer coefficients are called irreducible over the integers or prime. The goal in factoring a polynomial is to use one or more factoring techniques until each of the polynomial’s factors is prime or irreducible. In this situation, the polynomial is said to be factored completely. Factoring

3. H.G/1100/063 In any factoring problem, the first step is to look for the greatest common factor. The greatest common factor is a n expression of the highest degree that divides each term of the polynomial. The distributive property in the reverse direction ab + ac = a(b + c) can be used to factor out the greatest common factor. Common Factors

4. H.G/1100/064 Factor: a. 18 x 3 + 27 x 2 b. x 2 ( x + 3) + 5( x + 3) Solution a. We begin by determining the greatest common factor. 9 is the greatest integer that divides 18 and 27. Furthermore, x 2 is the greatest expression that divides x 3 and x 2. Thus, the greatest common factor of the two terms in the polynomial is 9x 2. 18x 3 + 27x 2 = 9x 2 (2x) + 9x 2 (3) Express each term with the greatest common factor as a factor. = 9x 2 (2 x + 3) Factor out the greatest common factor. b. In this situation, the greatest common factor is the common binomial factor (x + 3). We factor out this common factor as follows. x 2 (x + 3) + 5(x + 3) = (x + 3)(x 2 + 5) Factor out the common binomial factor. Text Example

5. H.G/1100/065 A Strategy for Factoring ax 2 + bx + c If no such combinations exist, the polynomial is prime. (Assume, for the moment, that there is no greatest common factor.) 1. Find two First terms whose product is ax 2 : ( x + )( x + ) = ax 2 + bx + c 2. Find two Last terms whose product is c: (x + )(x + ) = ax 2 + bx + c I 3. By trial and error, perform steps 1 and 2 until the sum of the Outside product and Inside product is bx: ( x + )( x + ) = ax 2 + bx + c O (sum of O + I)

6. H.G/1100/066 Factor: a. x 2 + 6 x + 8b. x 2 + 3 x – 18 Solution a. The factors of the first term are x and x: (x )( x ) To find the second term of each factor, we must find two numbers whose product is 8 and whose sum is 6. -6-969Sum of Factors -4, -2-8, -14, 28, 1Factors of 8 From the table above, we see that 4 and 2 are the required integers. Thus, x 2 + 6x + 8 = (x + 4)( x + 2) or (x + 2)( x + 4). This is the desired sum. Text Example

7. H.G/1100/067 Factor: a. x 2 + 6 x + 8b. x 2 + 3 x – 18 Solution b. We begin with x 2 + 3x – 18 = (x )( x ). To find the second term of each factor, we must find two numbers whose product is –18 and whose sum is 3. From the table above, we see that 6 and –3 are the required integers. Thus, x 2 + 3x – 18 = (x + 6)(x – 3) or (x – 3)(x + 6). -33-77-1717Sum of Factors -6, 36, -3-9, 29, -2-18, 118, -1Factors of -18 This is the desired sum. Text Example cont.

8. H.G/1100/068 Factor: 8 x 2 – 10 x – 3. Step 2Find two Last terms whose product is –3. The possible factors are 1(-3) and –1(3). Step 3Try various combinations of these factors. The correct factorization of 8x 2 – 10x – 3 is the one in which the sum of the Outside and Inside products is equal to –10x. Here is a list of possible factors. Solution Step 1Find two First terms whose product is 8x 2. 8x 2 – 10x – 3 (8x )(x ) 8x 2 – 10x – 3 (4x )(2x ) Text Example

9. H.G/1100/069 -4x + 6x = 2x(4x + 3)(2x – 1) 12x – 2x = 10x(4x – 1)(2x + 3) 4x – 6x = -2x(4x – 3)(2x + 1) -12x + 2x = -10x(4x + 1)(2x – 3) -8x + 3x = -5x(8x + 3)(x – 1) 24x – x = 23x(8x – 1)(x +3) 8x – 3x = 5x(8x – 3)(x + 1) -24x + x = -23x(8x + 1)(x – 3) Sum of Outside and Inside Products (Should Equal –10x) Possible Factors of 8x 2 – 10x – 3 Thus, 8x 2 – 10x – 3 = (4x + 1)(2x – 3) or (2x – 3)(4x + 1). This is the required middle term. Text Example cont.

10. H.G/1100/0610 The Difference of Two Squares If A and B are real numbers, variables, or algebraic expressions, then A 2 – B 2 = ( A + B )( A – B ). In words: The difference of the squares of two terms factors as the product of a sum and the difference of those terms.

11. H.G/1100/0611 Text Example v Factor: 81x 2 - 49 Solution: 81x 2 – 49 = (9x) 2 – 7 2 = (9x + 7)(9x – 7).

12. H.G/1100/0612 Factoring Perfect Square Trinomials Let A and B be real numbers, variables, or algebraic expressions, 1. A 2 + 2AB + B 2 = (A + B) 2 2. A 2 – 2AB + B 2 = (A – B) 2 v Here’s how to recognize a perfect square trinomial v 1. The first and last terms are squares of monomials or integers. v 2. The middle term is twice the product of the expressions being squared in the first and last terms.

13. H.G/1100/0613 Text Example v Factor: x 2 + 6 x + 9. x 2 + 6x + 9 = x 2 + 2 · x · 3 + 3 2 = (x + 3) 2 Solution:

14. H.G/1100/0614 Text Example v Factor: 25 x 2 – 60 x + 36. Solution: 25x 2 – 60x + 36 = (5x) 2 – 2 · 5x · 6 + 6 2 = (5x -6) 2.

15. H.G/1100/0615 Factoring the Sum and Difference of 2 Cubes 64x 3 – 125 = (4x) 3 – 5 3 = (4x – 5)(4x) 2 + (4x)(5) + 5 2 ) = (4x – 5)(16x 2 + 20x + 25) A 3 – B 3 = (A – B)(A 2 + 2AB + B 2 ) x 3 + 8 = x 3 + 2 3 = (x + 2)( x 2 – x·2 + 2 2 ) = (x + 2)( x 2 – 2x + 4) A 3 + B 3 = (A + B)(A 2 – 2AB + B 2 ) ExampleType

16. H.G/1100/0616 A Strategy for Factoring a Polynomial 1. If there is a common factor, factor out the GCF. 2. Determine the number of terms in the polynomial and try factoring as follows: a)If there are two terms, can the binomial be factored by one of the special forms including difference of two squares, sum of two cubes, or difference of two cubes? b)If there are three terms, is the trinomial a perfect square trinomial? If the trinomial is not a perfects square trinomial, try factoring by trial and error. c)If there are four or more terms, try factoring by grouping. 3. Check to see if any factors with more than one term in the factored polynomial can be factored further. If so, factor completely.

17. H.G/1100/0617 Factor: x 3 – 5 x 2 – 4 x + 20 Solution x 3 – 5x 2 – 4x + 20 = (x 3 – 5x 2 ) + (-4x + 20) Group the terms with common factors. = x 2 (x – 5) – 4(x – 5) Factor from each group. = (x – 5)(x 2 – 4) Factor out the common binomial factor, (x – 5). = (x – 5)(x + 2)(x – 2) Factor completely by factoring x 2 – 4 as the difference of two squares. Example

18. H.G/1100/0618 Rational Expressions http://www.nccu.edu/artsci/math/Gevorgyan

19. H.G/1100/0619 Example v If x 2 – 7 = 28, what is the value of x 2 + 7 = v If 4x – 5y = 15 and 2x – y = 9 then 6x – 6y =

20. H.G/1100/0620 Rational expressions v A rational expression is the quotient of two polynomials. v The set of real numbers for which an algebraic expression is defined is the domain of the expression. Because rational expression indicate division and division by zero is undefined, we must exclude numbers from a rational expression’s domain that make the denominator zero.

21. H.G/1100/0621 Find all the numbers that must be excluded from the domain of each rational expression. This denominator would equal zero if x = 2. This denominator would equal zero if x = -1. This denominator would equal zero if x = 1. SolutionTo determine the numbers that must be excluded from each domain, examine the denominators. Text Example

22. H.G/1100/0622 Simplifying Rational Expressions 1. Factor the numerator and denominator completely. 1. Divide both the numerator and denominator by the common factors.

23. H.G/1100/0623 Example v Simplify: Solution:

24. H.G/1100/0624 Multiplying Rational Expressions 1. Factoring all numerators and denominators completely. 2. Dividing both the numerator and denominator by common factors. 3. Multiply the remaining factors in the numerator and multiply the remaining factors in the denominator.

25. H.G/1100/0625 Example v Multiply and simplify: Solution:

26. H.G/1100/0626 Example v Divide and simplify: Solution:

27. H.G/1100/0627 Example v Add: Solution:

28. H.G/1100/0628 Finding the Least Common Denominator 1. Factor each denominator completely. 2. List the factors of the first denominator. 3. Add to the list in step 2 any factors of the second denominator that do not appear in the list. 4. Form the product of each different factor from the list in step 3. This product is the least common denominator.

29. H.G/1100/0629 Adding and Subtracting Rational Expressions That Have Different Denominators with Shared Factors 1. Find the least common denominator. 2. Write all rational expressions in terms of the least common denominator. To do so, multiply both the numerator and the denominator of each rational expression by any factor(s) needed to convert the denominator into the least common denominator. 3. Add or subtract the numerators, placing the resulting expression over the least common denominator. 4. If necessary, simplify the resulting rational expression.

30. H.G/1100/0630 Example v Subtract: Solution: